# POTWE-13-AE-23-S

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WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionThese are Sum Boxes Problem In a sequence of 12 numbers, each number after the ﬁrst three is equal to thesum of the previous three numbers. The 3 rd number in the sequence is 6, the6 th number in the sequence is 11, and the 11 th number in the sequence is 14.Determine all 12 numbers in the sequence. Solution Let  a 1  be the ﬁrst number in the sequence,  a 2  be the second,  a 4  be the fourth,and so on, until  a 12  which is the 12 th number in the sequence. The 12 boxes arelabelled in the following diagram. 6 11 14a 1  a 2  a 4  a 5  a 7  a 8  a 9  a 10  a 12 Each number after the third number is equal to the sum of the previous threenumbers. Therefore, looking at the 6 th term, we have 11 = 6 +  a 4  +  a 5  or a 4  +  a 5  = 5.Looking at the 7 th term,  a 7  =  a 4  +  a 5  + 11 = 5 + 11 = 16, since  a 4  +  a 5  = 5. 6 11 14a 1  a 2  a 4  a 5  16 a 8  a 9  a 10  a 12 Looking at the 9 th term,  a 9  = 11 + 16 +  a 8  = 27 +  a 8 .Looking at the 10 th term,  a 10  = 16 +  a 8  +  a 9  = 16 + ( a 8 ) +( a 8  +27) = 2 a 8  +43.Looking at the 11 th term, a 11  =  a 8  +  a 9  +  a 10  = ( a 8 ) + ( a 8  + 27) + (2 a 8  + 43) = 4 a 8  + 70.We are given that the 11 th term is  a 11  = 14.Therefore, 4 a 8  + 70 = 14, or 4 a 8  = − 56, or  a 8  = − 14.Therefore,  a 9  =  a 8  + 27 = − 14 + 27 = 13. a 10  = 2 a 8  + 43 = 2( − 14) + 43 = 15.Also,  a 12  =  a 9  +  a 10  +  a 11  = 13 + 15 + 14 = 42.  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING So far, we know the sequence is 6 11 14a 1  a 2  a 4  a 5  16 -14 13 15 42 Working backwards,  − 14 =  a 5  + 11 + 16, so  a 5  = − 41.From earlier,  a 4  +  a 5  = 5 and  a 5  = − 41, so  a 4  = 46.Continuing backwards,  a 5  =  a 2  + 6 +  a 4 , so  − 41 =  a 2  + 6 + 46, or  a 2  = − 93.And ﬁnally,  a 4  =  a 1  +  a 2  + 6, so 46 =  a 1  + ( − 93) + 6, or  a 1  = 133.Therefore, the sequence of 12 numbers is133 ,  − 93 ,  6 ,  46 ,  − 41 ,  11 ,  16 ,  − 14 ,  13 ,  15 ,  14 ,  42 6 11 14133-93 46 16 -14 13 15 42-41 We can indeed check that in this sequence, each number after the ﬁrst threenumbers is equal to the sum of the previous three numbers.

Jul 23, 2017

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Jul 23, 2017
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